The alias method, as stated in Sec. 8.4.3, requires generating at least two U(0, 1) random numbers—one to generate I in step 1 and the other to determine whether I or its alias is returned in step 2. Show that the following version of the alias method, which requires only one random number, is also valid:
1. Generate U ~ U(0, 1).
2. Let V = (n + 1)U, I = ⌊V⌋, and Uʹ = V – I.
3. If Uʹ ≤ FI, return X = I. Otherwise, return X = LI.
[Hint: What is the joint distribution of I and Uʹ? Although this “trick” does reduce the number of random numbers generated, it is probably not a good idea, since it depends on the low-order (least significant) bits of V – I being “random,” which may be doubtful for many (pseudo) random-number generators.]
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